|dw:1359238717297:dw|They told us that \(h=3r\),
which we can rewrite as \(\dfrac{1}{3}h=r\).
From here, we can substitute this \(h\) value in place of all the \(r\)'s in the formula.
\[\large A=2\pi r^2+2\pi r h \qquad \rightarrow \qquad A=2\pi\left(\dfrac{1}{3}h\right)^2+2\pi\left(\dfrac{1}{3}h\right)h\]

\[\large 402=2\pi \color{orangered}{r}^2+2\pi \color{orangered}{r} h \qquad \rightarrow \qquad 402=2\pi\color{orangered}{\left(\dfrac{1}{3}h\right)}^2+2\pi\color{orangered}{\left(\dfrac{1}{3}h\right)}h\]So we found or relationship with r and h and made the replacement.
Now let's try to solve for h.\[\large 402=2\pi\left(\dfrac{1}{3}h\right)^2+\color{royalblue}{2\pi\left(\dfrac{1}{3}h\right)h}\]Let's work on the blue part first.
\[\large 402=2\pi\left(\dfrac{1}{3}h\right)^2+\color{royalblue}{\dfrac{2\pi}{3}h^2}\]

\[\large 402=\color{#CC0033}{2\pi\left(\dfrac{1}{3}h\right)^2}+\dfrac{2\pi}{3}h^2\]Now to simplify the red term, make sure you square both the h and the 1/3.\[\large 402=\color{#CC0033}{2\pi\left(\dfrac{1}{9}h^2\right)}+\dfrac{2\pi}{3}h^2\]Which simplifies to,\[\large 402=\color{#CC0033}{\dfrac{2\pi}{9}h^2}+\dfrac{2\pi}{3}h^2\]

We have a couple of fractions, let's get a common denominator, multiplying the second term by \(\dfrac{3}{3}\).\[\large 402=\dfrac{2\pi}{9}h^2+\color{#996666}{\dfrac{3}{3}}\cdot\dfrac{2\pi}{3}h^2\]Giving us,\[\large 402=\dfrac{2\pi}{9}h^2+\dfrac{6\pi}{9}h^2\]Adding these terms together gives us,\[\large 402=\dfrac{8\pi}{9}h^2\]

To get rid of the fraction on the right, we'll multiply both sides by it's reciprocal.\[\large \color{#662FFF}{\left(\dfrac{9}{8\pi}\right)}402=\dfrac{8\pi}{9}h^2\color{#662FFF}{\left(\dfrac{9}{8\pi}\right)}\]We can cancel the fractions on the right,\[\large \color{#662FFF}{\left(\dfrac{9}{8\pi}\right)}402=\cancel{\dfrac{8\pi}{9}}h^2\cancel{\color{#662FFF}{\left(\dfrac{9}{8\pi}\right)}}\]Giving us,\[\large h^2=\dfrac{9\cdot402}{8\pi}\]